Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = 6.304 x 10^-3 in.) can be produced from the copper in 6.95 lb of covellite, an ore of copper is 66.0% copper by mass? (Hint: Treat the wire as a cylinder: V for cylinder = (pi)(r^2)(h); d of copper = 8.95 g/cm^3)

Answer: The atomic mass of E is 16

Explanation:Please see attachment for explanation

Answer:

The molar mass of E = 15.96 g/mol

Explanation:

Step 1: Data given

The compound Cr2E3 contains 68.47 % Cr and 31.53 % E

The molar mass of Cr = 52 g/mol

⇒ 2*52 = 104 g/mol

The molar mass of Cr2E3 = 2*52 g/mol + 3X

Step 2: Calculate mass of Cr2E3

Cr is 68.47 %

Mass of Cr = 104 grams

Cr2E3 is 100%

Mass of Cr2E3 = 151.89 grams

Molar mass of Cr2E3 = 151.89 g/mol

Step 3: Calculate mass of E

Mass of E3 = mass of Cr2E3 - mass of Cr

Mass of E3 = 151.89 grams - 104 grams

Mass of E3= 47.89 grams

Mass of E = 15.96 grams

The molar mass of E = 15.96 g/mol

Answer:

In the arrangement of particles within any atom, the outermost sort of particle is always the electron.

Explanation:

An atom consist of electron, protons and neutrons. Protons and neutrons are present with in nucleus while the electrons are present out side the nucleus.

All these three subatomic particles construct an atom. A neutral atom have equal number of proton and electron. In other words we can say that negative and positive charges are equal in magnitude and cancel the each other.

The electron is subatomic particle that revolve around outside the nucleus and has negligible mass. It has a negative charge.

Symbol= e-

Mass= 9.10938356×10-31 Kg

It was discovered by j. j. Thomson in 1897 during the study of cathode ray properties.

While neutron and proton are present inside the nucleus. Proton has positive charge while neutron is electrically neutral. Proton is discovered by Rutherford while neutron is discovered by James Chadwick in 1932.

Symbol of proton= P+  

Symbol of neutron= n0  

Mass of proton=1.672623×10-27 Kg

Mass of neutron=1.674929×10-27 Kg

In any atom, the outermost type of particle is the electron, specifically the valence electrons in the outermost shell. These electrons play a significant role in chemical reactions and bonding properties of the element. The organization of the Periodic Table reflects patterns in valence electron configurations that correspond to chemical behaviors.

In the arrangement of particles within any atom, the outermost sort of particle is always the electron. Electrons are the smallest of the three types of sub-atomic particles, carrying a negative charge and occupying the space outside the atomic nucleus. Inside the nucleus, much larger particles—protons and neutrons—are found, with protons having a positive charge and neutrons being electrically neutral.

The outermost electrons are of particular importance because they are the valence electrons. These are the electrons that reside in the outermost shell, or valence shell, of an atom in its uncombined state, and they play critical roles in determining the chemical properties of an element as well as its ability to form bonds with other atoms. The electron configuration of an atom is notably important because atoms with the same outer electron configurations tend to show similar chemical behavior, as demonstrated in the organization of the Periodic Table.

The number of valence electrons in the outermost shell can define an element's chemical reactivity and the types of bonds it can form. The arrangement of electrons in atoms means that the electrons with the highest energy levels, which are the valence electrons, are more likely to interact in chemical reactions than the core electrons which are closer to the nucleus and have lower energy levels.

Answer: The vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = initial pressure at [tex]25^oC[/tex] = 23.8 torr

[tex]P_2[/tex] = final pressure at [tex]50^oC[/tex] = ?

[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 43.9 kJ/mol = 43900 J/mol

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]

[tex]T_2[/tex] = final temperature = [tex]50^oC=273+50=323K[/tex]

Now put all the given values in this formula, we get

[tex]\log (\frac{P_2}{23.8}=\frac{43900}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{323K}][/tex]

[tex]\frac{P_2}{23.8}=antilog(0.5955)[/tex]

[tex]P_2=93.8torr[/tex]

Therefore, the vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr

Answer:

0.0016 mol/(L.s)

Explanation:

The rate of a reaction (r) can be calculated by the initial concentration of the reagent, by the expression:

-r = k*[reagent]ⁿ

Where the minus sign represents that the reagent is disappearing, k is the rate constant, which depends on the temperature, and n is the order of the reaction. For the reaction with more than 1 reagent, each reagent will have its order, which is determined by experiments. So, for n = 0:

-r = 0.0016*(1.50)⁰

-r = 0.0016 mol/(L.s)

Final answer:

The rate of a zero-order reaction is equal to the rate constant, which in the case of the hydrogenation of ethylene using a nickel catalyst is 0.0016 mol L.

Explanation:

To determine the rate of reaction for a zero-order reaction, we use the rate law which states that the rate is independent of the concentration of the reactants. Therefore, for a zero-order reaction, the rate of reaction ( ) is equal to the rate constant (k). Given that the rate constant k is 0.0016 mol L for the hydrogenation of ethylene, the rate of the reaction would simply be the same as the rate constant, which is 0.0016 mol L

The question is incomplete, here is the complete question.

A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride [tex]CuF_2[/tex] solution to a reaction flask.

Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

Answer : The mass of copper(II) fluoride is, 0.13 mg

Explanation :  Given,

Millimolarity of copper (II) fluoride = 0.0013 mM

This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution

Converting millimoles into moles, we use the conversion factor:

1 moles = 1000 millimoles

So, [tex]0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol[/tex]

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

We are given:

Moles of copper (II) fluoride solution = [tex]1.3\times 10^{-6}mol[/tex]

Molar mass of copper (II) fluoride = 101.5 g/mol

Putting values in above equation, we get:

[tex]1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g[/tex]

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

[tex]\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg[/tex]

Therefore, the mass of copper(II) fluoride is, 0.13 mg

Answer:

the mass of copper(II) fluoride added to the flask is approximately 476,250 micrograms.

Explanation:

To calculate the mass of copper(II) fluoride (CuF2) added to the reaction flask, you need to multiply the volume (in liters) of the solution by its molarity and then convert the result to micrograms.

Given:

Volume of the solution (V) = 0.015 L

Molarity of the solution (M) = 0.500 M

First, calculate the number of moles of copper(II) fluoride using the formula:

Moles (n) = Molarity (M) × Volume (V)

n = 0.500 M × 0.015 L = 0.0075 moles

Now, you need to convert moles to micrograms. One mole of any substance contains Avogadro's number of molecules (6.022 × 10^23). You also need to consider the molar mass of copper(II) fluoride (CuF2), which is the sum of the atomic masses of copper (Cu) and two fluorine (F) atoms:

Molar mass of CuF2 = (1 * Cu) + (2 * F) = 63.5 g/mol (approximately)

Now, calculate the mass in grams:

Mass (g) = Moles (n) × Molar mass (Molar mass of CuF2)

Mass (g) = 0.0075 moles × 63.5 g/mol ≈ 0.47625 g

Now, convert grams to micrograms. There are 1,000,000 micrograms in a gram:

Mass (micrograms) = Mass (g) × 1,000,000

Mass (micrograms) = 0.47625 g × 1,000,000 µg/g = 476,250 µg

So, the mass of copper(II) fluoride added to the flask is approximately 476,250 micrograms.

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Answer:

Explanation:Isobaric is a thermodynamic process in which the pressure of the system is zero. It is a process where there is no work done in the system. Therefore the temperature of the system according to the ideal gas law will vary linearly with pressure and inversely with volume.

Answer : A nucleotide is composed of a phosphate group and a nitrogen-containing base.

Explanation :

Nucleotide : It is a building block of nucleic acids or we can say that it is building block of DNA and RNA.

It is composed of three sub-unit molecules which are a nitrogenous base, a five-carbon sugar and one phosphate group.

Nucleotide forms covalent bonds with other nucleotide for the formation of the nucleic acid strand.

Hence, a nucleotide is composed of a phosphate group and a nitrogen-containing base.

A nucleotide is composed of a a. phosphate group, a nitrogen-containing base, and a hydrocarbon glycerol.

A nucleotide is a fundamental building block of nucleic acids, such as DNA and RNA. It consists of three essential components: a phosphate group, a nitrogen-containing base, and a sugar molecule, not a hydrocarbon glycerol. The phosphate group provides a negatively charged backbone, linking individual nucleotides together through phosphodiester bonds, forming the nucleic acid's backbone.

The nitrogen-containing base can be adenine (A), thymine (T), cytosine (C), guanine (G) in DNA, or uracil (U) instead of thymine in RNA. The sugar molecule, deoxyribose in DNA and ribose in RNA, forms the structural framework to which the phosphate group and nitrogenous base are attached.

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The volume of a potassium iodide solution with a concentration of 0.0380 M and containing 150 g of potassium iodide is 3.95 L, rounded to three significant digits.

To calculate the volume V of a solution, you can use the formula:

[tex]\[ V = \frac{\text{mass of solute}}{\text{molarity}} \][/tex]

Given that the molarity M is [tex]\(0.0380 \, \text{M}\)[/tex] and the mass of potassium iodide is [tex]\(150 \, \text{g}\)[/tex], substitute these values into the formula:

[tex]\[ V = \frac{150 \, \text{g}}{0.0380 \, \text{M}} \]\[ V = 3947 \, \text{mL} \][/tex]

Since the answer should have the correct number of significant digits, the volume is [tex]\(3.95 \, \text{L}\)[/tex] (rounded to three significant digits).

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The complete question is:

Calculate the volume in liters of a 0.0380M potassium iodide solution that contains 150 g of potassium iodide. Be sure your answer has the correct number of significant digits.

Answer:

C21H30O5

Explanation:

Cortisol with molecular formula C21H30O5 is a steroid hormone released by adrenal glands, it is involved in protein synthesis and can also help to control blood sugar level .

Answer:

the transition state will resemble the products more than the reactants

Explanation:

Since the free-radical bromination of methane involves the following reactions

Br₂ → 2 Br•  , ΔH⁰ (per mole) = +192 kJ , Ea (per mole) = + 192 kJ

CH₄ +Br•   → CH₃• + HBr  , ΔH⁰ (per mole) = +67 kJ , Ea (per mole) = + 75 kJ

CH₃• +Br₂  → CH₃Br  + Br•  , ΔH⁰ (per mole) = -101 kJ , Ea (per mole) = -4 kJ

without considering the Br dissociation (initiation reaction) the rate-determining step is the second equation ( highest Ea) .

Then since the reaction is endothermic (ΔH⁰ (per mole) = +67 kJ) , the transition state will resemble the products more than the reactants ( Hammond postulate)

The free-radical bromination of methane's transition state resembles both the products and reactants equally. This similarity arises due to the formation and subsequent decay of the high-energy transition state that exists between reactants and products. The reaction's activation energy is always positive and its exothermic nature results in a decrease in system enthalpy.

In the free-radical bromination of methane, the nature of the transition state of the rate-determining step is closest to option b. The transition state resembles both the products and reactants equally. This is because during the process, reactant molecules with enough energy collide to form a high-energy activated complex or transition state. This transition state then decays to yield stable products, maintaining similarities to both the initial reactants and final products.

As per chemical kinetics, the reaction diagram indicating this process highlights the activation energy, Eå, as the energy difference between the reactants and the transition state. The enthalpy change of the reaction, ΔH, is evaluated as the energy difference between the reactants and products. Here, the reaction is exothermic (ΔH < 0) as it results in a decrease in system enthalpy.

Understanding that the activation energy is always positive in these reactions, regardless of whether the reaction is exergonic (releases energy) or endergonic (absorbs energy), further supports the resemblance between the transition state and both reactants and products.

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Answer: 2,5-dimethyl octane.

Explanation:

The basic rules for naming of organic compounds are :

1. First select the longest possible carbon chain.

2. The longest possible carbon chain should include the carbons of double or triple bonds.

3. The naming of alkane is done by adding the suffix -ane, alkene by adding the suffix -ene, alkyne by adding the suffix -yne and carboxylic acid by adding the suffix -oic acid.

4. The numbering is done in such a way that first carbon of double or triple bond gets the lowest number.

5. The carbon atoms of the double or triple bond get the preference over the other substituents present in the parent chain.

6. If two or more similar alkyl groups are present in a compound, the words di-, tri-, tetra- and so on are used to specify the number of times of the alkyl groups in the chain.

Thus the name for straight chain made of 8 carbons with [tex]-CH_3[/tex] groups on the second and fifth carbons will be 2,5-dimethyl octane.

The desired compound using a stepwise approach from dicyclopentadiene and incorporating a Diels-Alder reaction .

Step 1: Diels-Alder Reaction

In the Diels-Alder reaction, dicyclopentadiene (DCPD) will react with a suitable dienophile to form the desired cycloadduct. The dienophile you choose should have a functional group that can be further manipulated to achieve the final compound. Let's select maleic anhydride (C4H2O3) as the dienophile, which can react with DCPD to yield a cycloadduct.

The reaction would look like this:

Dicyclopentadiene + Maleic Anhydride → Cycloadduct

Step 2: Functionalization of the Cycloadduct

The cycloadduct formed in the Diels-Alder reaction will contain the necessary functional groups for further manipulation. In this case, you want to introduce additional substituents or functional groups.

For example, you can introduce an alcohol group (-OH) through a nucleophilic addition reaction. You can do this by treating the cycloadduct with a strong base (e.g., sodium hydroxide, NaOH) and water (H2O) to open the anhydride ring and form a carboxylic acid intermediate. Then, you can reduce the carboxylic acid using a reducing agent (e.g., lithium aluminum hydride, LiAlH4) to obtain the alcohol group.

The reaction sequence would look like this:

Cycloadduct + NaOH + H2O → Carboxylic Acid Intermediate

Carboxylic Acid Intermediate + LiAlH4 → Alcohol Group

Step 3: Further Functionalization

Depending on your specific requirements, you can further modify the alcohol group by various organic reactions such as esterification, acylation, or oxidation, among others, to achieve the desired final compound.

This stepwise synthesis should allow you to obtain the desired compound from dicyclopentadiene, using a Diels-Alder reaction as one of the key steps. Keep in mind that the reaction conditions and reagents may need to be optimized based on the specific compound you are targeting. Always follow safety guidelines and consult relevant chemical literature for detailed reaction conditions.

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To synthesize the given compound from dicyclopentadiene, a stepwise approach involving a Diels-Alder reaction can be followed. First, dicyclopentadiene is converted to cyclopentadiene. Then, cyclopentadiene reacts with maleic anhydride to form the Diels-Alder adduct.

A stepwise synthesis of the given compound from dicyclopentadiene using a Diels-Alder reaction as one step can be achieved as follows:

Note that the specific reagents and conditions for each step may vary depending on the desired compound and the available starting materials. Finally, the adduct can be further modified to obtain the desired compound.

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Answer:

2H2S + 3O2 → 2SO2 + 2H2O

Explanation:

Step 1: Data given

Hydrogen sulfide = H2S

Oxygen = O2

sulfur dioxide = SO2

water = H2O

Step 2: The unbalanced equation

H2S + O2 → SO2 + H2O

Step 3: Balancing the equation

H2S + O2 → SO2 + H2O

On the left side we have 2x O (in O2) and on the right side we have 3x O (2x in SO2 and 1x in H2O). To balance the amount of O, we have to multiply O2 (on the left side) by 3 and SO2 and H2O on the right side by 3.

H2S + 3O2 → 2SO2 + 2H2O

On the right side we have 4x H and on the left side we have 2x H. To balance the amount of H, we have to multiply H2S by 2.

Now the equation is balanced.

2H2S + 3O2 → 2SO2 + 2H2O

Answer:

4. Oxygen

Explanation:

Answer:

The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.

Explanation:

To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.

It is given by Bohr's Theory as:

1/λ = Rh (1/n1² - 1/n2²)

where,

λ = wavelength of photon

n1 = initial state = 1 (ground-state of hydrogen)

n2 = final state = ∞ (since, electron goes far away from atom after ionization)

Rh = Rhydberg's Constant = 1.097 x 10^7 /m

Therefore,

1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)

λ = 9.115 x 10^-8 m = 91.15 nm

Now, for energy (E) we know that:

E = hc/λ

where,

h = Plank's Constant = 6.625 x 10^-34 J.s

c = speed of light = 3 x 10^8 m/s

Therefore,

E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)

E = 2.18 x 10^-18 J

E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)

E = 13.6 eV

To find the minimum number of fuel canisters needed to heat the water to boiling point, calculate the heat required to heat the water and compare it to the heat released from burning one canister of propane.

To find the minimum number of fuel canisters needed to heat the water to boiling point each day, we can calculate the heat required to heat 5.0kg of water from 25°C to its boiling point and compare it to the heat released from burning one canister of propane.

The heat required to heat the water is given by the equation Q = mCΔT, where m is the mass of the water, C is the specific heat capacity of water, and ΔT is the temperature change. The specific heat capacity of water is approximately 4.184 J/g°C.

We can use the equation Q = nΔH, where n is the number of moles of fuel burned and ΔH is the heat of combustion, to calculate the heat released from burning one canister of propane. The heat of combustion of propane is given as -2219.2 kJ/mol.

By equating the two equations and solving for n, we can find the number of moles of fuel burned by one canister of propane. Then, we can use the molar mass of propane (44.1 g/mol) to find the mass of propane burned by one canister. Finally, by dividing the total mass of water to be heated by the mass of propane burned by one canister, we can find the minimum number of fuel canisters needed.

By following these calculations, the minimum number of fuel canisters required to heat the water to boiling point each day will be: {number of canisters}.

Answer: the empirical formula is C3H4O3

Explanation:Please see attachment for explanation

Answer:

The answer to your question is  C₃H₄O₃

Explanation:

Data

CxHyOz

mass of sample = 0.150 g

mass of CO₂ = 0.225 g

mass of H₂O = 0.0614 g

Reaction

                     CxHyOz + O₂   ⇒   CO₂   + H₂O

Process

1.- Calculate the moles of C

                            44 g of CO₂ ----------------- 12 g of C

                            0.225 g    ---------------- x

                            x = (0.225 x 12) / 44

                            x = 0.0614 g of C

                            12 g of C -------------------- 1 mol

                           0.0614 g of C --------------- x

                                 x = 0.0051 moles of Carbon

2.- Calculate the moles of hydrogen

                            18 g of H₂O ------------------ 2 g of H

                             0.0614 g --------------- x

                             x = 0.0068 g of H

                             1 g of H ----------------------- 1 mol of H

                            0.0068 g --------------------- x

                             x = 0.0068 moles of H

3.- Calculate the mass of Oxygen

    Mass of oxygen = 0.150 - 0.0614 - 0.0068

                               = 0.0818 g

                       16 g of O -------------------  1 mol

                      0.0818 g -------------------- x

                         x = (0.0818 x 1) / 16

                         x = 0.0051 moles of O

4.- Divide by the lowest number of moles

Carbon       0.0051 / 0.0051  = 1

Hydrogen   0.0068 / 0.0051 = 1.33

Oxygen       0.0051 / 0.0051 = 1

Multiply these numbers by 3

Carbon 3

Hydrogen = 4

Oxygen = 3

5.- Write the empirical formula

                                      C₃H₄O₃

Q: What is the change of entropy for 3.0 kg of water when the 3.0 kg of water is changed to ice at 0 °C? (Lf = 3.34 x 105 J/kg)

Answer:

-3670.33 J/K

Explanation:

Entropy: This can be defined as the degree of randomness or disorderliness of a substance. The S.I unit of Entropy is J/K.

Mathematically,  change of Entropy can be expressed as,

ΔS = ΔH/T ....................................... Equation 1

Where ΔS = Change of entropy, ΔH = heat change, T = temperature.

ΔH = -(Lf×m).................................... Equation 2

Note: ΔH is negative because heat is lost.

Where Lf = latent heat of ice = 3.34×10⁵ J/kg, m = 3.0 kg, m = mass of water = 3.0 kg

Substitute into equation

ΔH = -(3.34×10⁵×3.0)

ΔH = - 1002000 J.

But T = 0 °C = (0+273) K = 273 K.

Substitute into equation 1

ΔS = -1002000/273

ΔS = -3670.33 J/K

Note: The negative value of ΔS shows that the entropy of water decreases when it is changed to ice at 0 °C

The change in entropy for 3.0 kg of water changing into ice at 0 degrees Celsius is approximately 3658 J/K. This is calculated using the formula for entropy change and the known values for the heat of fusion for water and the temperature in Kelvin.

The change in entropy during a phase transition, such as the one from water to ice, can be calculated using the formula ΔS = Q/T, where ΔS is the change in entropy, Q is the heat involved during the phase transition, and T is the temperature in Kelvin. For the given scenario, the phase change is from liquid to solid (freezing), and the heat involved (Q) is equal to the mass multiplied by the heat of fusion for water. The heat of fusion for water is 333.55 J/g and the temperature (T) at which this change occurs is 0 degrees Celsius or 273.15 Kelvin. The mass is 3.0 kg or 3000g. Therefore, Q = 3000g x 333.55J/g = 999150 J. Substituting these values in our formula, ΔS = 999150 J / 273.15 K ≈ 3658 J/K.

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Answer:

Hi

A conjugate acid is a chemical compound formed by the reception of a proton by a base; therefore, it is a base with an added hydrogen ion. On the other hand, a conjugate base is what remains after an acid has donated its proton during a chemical reaction, so it can be said that a conjugate base is a species modified by extracting a proton from an acid.

In pyrrole, the electron pair can be relocated to the ring and its availability is very small. In addition, the concept of aromaticity comes into play.

In the attached file are the schemas of the conjugate bases.

Explanation:

Answer:

mm protein = 15365.8183 g/mol

Explanation:

∴ molar mass (mm) ≡ g/mol

∴ π = 0.118 atm

∴ T = 25°C ≅ 298 K

∴ concentration (C ) [=] mol/L

∴ mass protein = 371 mg = 0.371 g

∴ volume sln = 5.00 mL = 5 E-3 L

⇒ C = π / RT = (0.118 atm)/((0.082 atm.L/K.mol)(298 K))

⇒ C = 4.8289 E-3 mol/L

⇒ mol protein = (4.8289 E-3 mol/L)×(5 E-3 L) = 2.4145 E-5 mol

⇒ mm protein = (0.371 g)/(2.4145 E-5 mol) = 15365.8183 g/mol

Final answer:

To find the molar mass of the protein, we begin by converting mass to grams and volume to liters, and then apply the van 't Hoff equation for osmotic pressure. After rearranging the equation, we calculate the molar mass using the given osmotic pressure, volume, temperature, and ideal gas constant.

Explanation:

To calculate the molar mass of the protein using the osmotic pressure, we can use the van 't Hoff equation for osmotic pressure, Π = MRT, where Π is the osmotic pressure, M is the molarity of the solution, R is the ideal gas constant, and T is the temperature in Kelvin. In this scenario, the student provided the mass of the protein (371 mg), the volume of the solution (5.00 mL), the osmotic pressure (0.118 atm), and the temperature at which the osmotic pressure was measured (25 °C which is 298.15 K).

First, convert the mass of the protein to grams:

371 mg = 0.371 g

Next, convert the volume from mL to liters:

5.00 mL = 0.005 L

Now, calculate the molarity (M) of the protein:

Number of moles = mass (g) / molar mass (g/mol)

As we do not know the molar mass yet, let's call it 'Mm'. The molarity (concentration) of the solution is:

Molarity (M) = Number of moles / Volume (L) = (mass (g) / Mm (g/mol)) / Volume (L)

Convert the temperature to Kelvin:

25 °C = 298.15 K

Using the van 't Hoff equation, we solve for the molar mass (Mm):

Π = (mass (g) / Mm (g/mol)) / Volume (L) × R × T

0.118 atm = (0.371 g / Mm (g/mol)) / 0.005 L × 0.0821 (L·atm/K·mol) × 298.15 K

Rearranging the equation and solving for Mm gives us:

Mm = (0.371 g / 0.005 L) / (0.118 atm / (0.0821 L·atm/K·mol × 298.15 K))

After calculating, we find the molar mass of the protein.

A

Individual lipid molecules are free to diffuse laterally in the surface of the bilayer is a general feature of the lipid bilayer in all biological membranes

Explanation:

The inside of the bilipid layer of the cell membrane of cells is made of fatty acid chains and cholesterol which are nonpolar molecules. they are sandwiched between the polar (glycerol) ends of the  chains because they are hydrophobic and cannot interact with the ‘watery’ extracellular fluids.  Non-polar molecules like lipids can easily diffuse laterally, within this lipid layers of the membrane because non-polar molecules interact well with other nonpolar molecules.

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Answer:

1) chemical change

2) filtration

Explanation:

A chemical change involves the formation of a new substance. In this case, the pure white solid formed is an entirely new substance, with a different chemical identity from those of the two solutions mixed to form it. The new solid is a precipitate. Precipitates are easily separated by filtration of the reaction mixture. Another name for chemical change is chemical reaction.

1) This is an example of chemical change.

2) The most convenient way to separate the newly formed white solid from the liquid mixture is filtration

A chemical change involves the formation of a new substance. In this case, the pure white solid formed is an entirely new substance, with a different chemical identity from those of the two solutions mixed to form it. The new solid is a precipitate. Precipitates are easily separated by filtration of the reaction mixture. Another name for chemical change is chemical reaction.

The most convenient way  to separate the newly formed white solid from the liquid mixture is filtration since precipitates are formed.

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Answer: 2.02mL

Explanation:

Density = 2.82 g/mL

Mass = 5.71 g

Volume =?

Density = Mass /volume

2.82 = 5.71 / Volume

Volume = 5.71 / 2.82

Volume = 2.02mL

The correct formula unit equation for the reaction of potassium sulfate and barium chloride forming potassium chloride and barium sulfate in water is: K2SO4 (aq) + BaCl2 (aq) → 2 KCl (aq) + BaSO4 (s). This is because both potassium sulfate and barium chloride are soluble in water, but barium sulfate is not.

The correct formula unit equation for the reaction of potassium sulfate and barium chloride forming potassium chloride and barium sulfate in water is: K2SO4 (aq) + BaCl2 (aq) → 2 KCl (aq) + BaSO4 (s).

This can be justified as when potassium sulfate (K2SO4) and barium chloride (BaCl2) are dissolved in water they dissociate into their respective ions, making them aqueous. The formation of potassium chloride (KCl) acknowledges that potassium ions and chloride ions can be attracted to each other in the water solution, so this is also aqueous. However, barium sulfate (BaSO4) is known to be insoluble in water, and hence, precipitates as a solid (s). Hence, option 2 is correct.

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Answer:

The answer is Inductive effect

Explanation:

To determine the acidity or alkalinity of an organic compound. We have to keep in mind that the whole analysis is based on the comparison between the compounds, and we must work with the conjugated base of the molecule. Keeping in mind, the more unstable the base, the less acidic the molecule is. Thus, to determine instability, the Inductive Effect of the molecule can be used.

This type of effect occurs when atoms of different electronegativities are linked or very close in the compound. The most electronegative atom has a tendency to bring electrons close to it, thus creating a dipole. This dipole can have a stabilizing effect on the molecule, as it “relieves” the excessive charge on some occasions, better accommodating the charges.

However, in some cases, instead of chains with chlorine radicals, we may have chains with methyl radicals. This has a major impact on the inductive effect, keeping in mind that alkyl groups are electron donors.

The question is incomplete, here is the complete question:

Consider this reaction occurring at 298 K:

[tex]N_2O(g)+NO_2(g)\rightleftharpoons 3NO(g)[/tex]

If a reaction mixture contains only [tex]N_2O\text{ and }NO_2[/tex] at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture.

What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous. Given that: [tex]\Delta G^o_{rxn}=107.8kJ/mol[/tex]

Answer: The maximum partial pressure of NO will be [tex]5.01\times 10^{-7}atm[/tex]

Explanation:

For the given chemical equation:

[tex]N_2O(g)+NO_2(g)\rightleftharpoons 3NO(g)[/tex]

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{NO}^3}{p_{N_2O}\times p_{NO_2}}[/tex]

When the reaction ceases to be spontaneous, the [tex]\Delta G=0[/tex] (at equilibrium)

Relation between standard Gibbs free energy and equilibrium constant follows:

[tex]\Delta G=\Delta G^o+2.303RT\log K_p[/tex]

where,

[tex]\Delta G^o[/tex] = Standard Gibbs free energy = 107.8 kJ/mol = 107800  J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = [tex]8.314J/K mol[/tex]

T = temperature = 298 K

[tex]p_{N_2O}=1.00atm[/tex]

[tex]p_{NO_2}=1.00atm[/tex]

Putting values in above equation, we get:

[tex]0=107800J/mol+(2.303\times 8.314J/Kmol)\times 298K\times \log (\frac{p_{NO}^3}{1.00\times 1.00})[/tex]

[tex]-107800=5705.85\times \log (\frac{p_{NO}^3}{1.00\times 1.00})\\\\-18.893=\log (p_{NO}^3)-\log (1.00)\\\\-18.893=3\log (p_{NO})\\\\\log (p_{NO})=-6.30\\\\p_{NO}=10^{-6.30}=5.01\times 10^{-7}atm[/tex]

Hence, the maximum partial pressure of NO will be [tex]5.01\times 10^{-7}atm[/tex]

Answer: (1). pH = 1.70

(2). pH = 2.3

(3). pH = 3.3

(4). pH = 4.3

(5). pH = 8.41

(6). pH = 10.22

Explanation:

we assume that the formula representation of acid is H₂A

the titration curve has reasonably sharp breaks at both equivalence points, corresponding to the reactions;

H₂A + OH⁻ → HA⁻ + H₂O

HA⁻ + OH⁻ → A²⁻ + H₂O

the volume of NaOH (V₀) at the first equivalent point is,

V₀ = (20.0 mL)(0.100M) / 0.100M = 20.0mL

where volume of NaOH at 1/2 equivalent point is,

V₀/2 = 10.0mL

also Volume of NaOH at the second equivalence (2V₀) point is 40.0mL

the volume of NaOH at 1/2 second equivalent point is,

V₀ + V₀/2 = 30.0mL

Volume of NaOH after second equivalence exceeds 40mL

therefore, at 0 mL NaOH addition;

where the extent of ionization is assumed to be x, we have

                        H₂A   ⇆     HA⁻   +   H⁺

where initial:   0.1 M       -            -

          change:   -x         +x           +x

          Equili:      0.1-x      x             x

Kаl = [HA⁻][H⁺] / [H₂A]

10⁻²³ = (x)(x) / (0.1-x)

x = 0.020

[H⁺] = 0.020 M

pH = -log [H⁺]

pH = -log(0.020)

pH = 1.70

(2). at 10 mL NaOH addition

[H₂A]ini = 0.10 M * 20.0 mL = 2 mmol

[OH⁻] = 0.1 M * 10 mL = 1 mmol

after reaction:

[H₂A] = 1 mmol

[H⁻] = 1 mmol

pH = pKa₁ + log [HA⁻] / [[HA⁻]

pH = 2.3 + log 1mmol / 1mmol

pH = 2.3

(3). pH at the first equivalence point is,

pH = 1/2 (pKa₁ + pKa₂)

pH = 1/2(2.3 + 4.3) = 3.3

pH = 3.3

(4). pH at the second 1/2 equivalence point is

pH = pKa₂ = 4.3

pH = 4.3

(5). pH at the second equivalence point;

all H₂A is converted into A²⁻

[A²⁻] = initial moles of H₂A / total volume = (20.0 mL)(0.10 M) / (20.0 + 40.0) mL = 0.033 M

at equilibrium:

                   A²⁻ + H²O    ⇆   HA⁺ OH⁻

          0.033 - x

from the Kb₁ expression,

Kb₁ = [OH⁻][HA⁻] / [A²]

Kw/Ka₂ = x²/(0.0333 - x)

10⁻¹⁴/10⁻⁴³ = x²/(0.0333 - x)

x = 2.57 * 10⁻⁶

[OH⁻] = 2.57 * 10⁻⁶M

pH = -log Kw/[OH⁻] = 8.41

pH = 8.41

(6). pH after second equivalence point;

assuming the volume of NaOH is 40.10 mL

after second equivalence point OH⁻ in excess

[OH⁻] = 0.10 M * 0.10 mL / (20 + 40.10) mL = 1.66 * 10⁻⁴ M

pH = 0=-log Kw/[OH⁻] = 10.22

pH = 10.22

Answer:

D is the true statement.

Explanation:

Recall that during a phase change the temperature remains constant until all the material has changed its phase. The energy put into the system is utilized to make the phase change, and afterward the temperature of the vliquid will start to increase.

In this case we have a mixture of liquid water and ice at  0ºC  ( assume standard pressure) which is the temperature in which liquid water and ice coexist. The ice will  melt until consumed at constant T = 0ºC, and then the temperature of the liquid water will start to increase at a uniform rate since we are heating at a constant rate.

Now we are in position to answer this question.

A- False the temperature does not increase as the ice melts.

B- False for the reasons given above.

C- False the temperature does not increase slowly as the ice melts, but remains constant.

D-True the temperature of the mixture does not change until all the ice has melted, and then the temperature of liquid water will start to increase uniformly.

Heat applied to a water-ice mixture initially goes into breaking the hydrogen bonds in ice (latent heat of fusion), without causing a temperature increase. Only after all the ice has melted, the added heat increases the water temperature.

The correct option is D: The temperature of the mixture does not change at all until all the ice has melted, at which point it increases at a constant rate. This is because the energy from the heat being applied is first used to break the hydrogen bonds in ice, a process known as latent heat of fusion, which does not involve an increase in temperature. Only after all the ice has melted will the added heat then lead to an increase in the temperature of the water.

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Answer:

a, b and d

Explanation:

Proper Solvent choice is very important for effective recrystallization.

Therefore, features of a good solvent are.

a. The solvent should dissolve a moderate quantity of the target substance near its boiling point but only.

b. The solvent should not react with the target substance.

d. The solvent should be easily removed from the purified product.

options c and d are not a property of good solvent.

Complete Question

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Answer:

This is shown on the second,third , fourth and fifth image

Explanation:

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